Find the maximum value a product

Question

Three positive integers their sum is 90, find the numbers for which their product is maximum. Let the numbers $x,y$ and $z$ we have $$x+y+z=90$$ Using AM-GM $$x+y+z\ge 3(xyz)^{\frac{1}{3}}$$ then $$xyz\le2700$$ then $$xyz=2700$$ Now we have two equations and 3 unknowns can we find those numbers using these two equations ?

Answer 1

HINT

Recall that for AM-GM equality holds for $x=y=z$.

Answer 2

You already have their maximum product, it's 27000. You know that they can achieve this value because the AM-GM has an equality case: when all variables are equal, i.e. $x=y=z$. So you are looking for three equal numbers $x,y,z$ adding up to 90, therefore they are $30$.

Answer 3

AM-GM inequality becomes equality if and only if $x=y=z$, so $3x=90$, $x=y=z=30$ and we really have $x+y+z=90=3\sqrt[3]{30\cdot 30\cdot 30}$.